Three Dimensional Geometry: Angles

Angle Between Two Lines

In three-dimensional space, two lines can be intersecting, parallel, or skew (neither intersecting nor parallel). The angle between two lines provides a measure of their relative orientation.

If two lines intersect, the angle between them is simply defined as the angle between their respective direction vectors. Since lines extend infinitely in both directions, there are typically two angles formed at the intersection: an acute angle and an obtuse angle (unless the lines are perpendicular or parallel). By convention, the angle between two lines is usually taken to be the acute angle.

If the two lines are parallel, the angle between them is defined as $0^\circ$.

If the two lines are skew (they do not intersect and are not parallel), the angle between them is defined as the angle between any two intersecting lines that are drawn parallel to the given skew lines. This angle is equal to the angle between the direction vectors of the two skew lines.

In all cases, the angle between two lines is the angle between their direction vectors.

Formula using Vector Form

Let the vector equations of two lines $L_1$ and $L_2$ be given by:

Here, $\vec{a_1}$ and $\vec{a_2}$ are the position vectors of points on lines $L_1$ and $L_2$ respectively. $\vec{b_1}$ is a vector parallel to line $L_1$ (its direction vector), and $\vec{b_2}$ is a vector parallel to line $L_2$ (its direction vector). $\lambda$ and $\mu$ are scalar parameters.

The angle $\theta$ between the lines $L_1$ and $L_2$ is the angle between their direction vectors $\vec{b_1}$ and $\vec{b_2}$. Using the definition of the dot product, we know that for any two non-zero vectors $\vec{u}$ and $\vec{v}$, $\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos \phi$, where $\phi$ is the angle between them. Solving for $\cos \phi$:

Applying this to the direction vectors $\vec{b_1}$ and $\vec{b_2}$, the angle $\phi$ between them is such that:

The angle $\theta$ between the lines is conventionally the acute angle, which means $0 \le \theta \le \frac{\pi}{2}$. The cosine of an acute angle is non-negative. Therefore, to find the acute angle $\theta$, we take the absolute value of the dot product in the numerator:

$$ \mathbf{\cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}} $$

where $0 \le \theta \le \frac{\pi}{2}$.

Formula using Cartesian Form

Let the Cartesian equations of two lines $L_1$ and $L_2$ be given by the symmetric form:

In these equations, $(a_1, b_1, c_1)$ are the direction ratios of line $L_1$, and $(a_2, b_2, c_2)$ are the direction ratios of line $L_2$.

A vector parallel to $L_1$ can be taken as $\vec{b_1} = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}$.

A vector parallel to $L_2$ can be taken as $\vec{b_2} = a_2\hat{i} + b_2\hat{j} + c_2\hat{k}$.

The angle $\theta$ between the lines is the acute angle between $\vec{b_1}$ and $\vec{b_2}$. We use the formula $\cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$.

Calculate the dot product $\vec{b_1} \cdot \vec{b_2}$ in component form:

Calculate the magnitudes $|\vec{b_1}|$ and $|\vec{b_2}|$:

Substitute these into the formula for $\cos \theta$:

$$ \mathbf{\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}} $$

This formula gives the cosine of the acute angle between two lines when their direction ratios are known.

If the direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ of the lines are known instead of direction ratios, the formula simplifies. Since $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ can be considered as the components of unit direction vectors $\hat{u_1}$ and $\hat{u_2}$ respectively (where $|\hat{u_1}|=1$ and $|\hat{u_2}|=1$), the formula becomes:

$$ \mathbf{\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|} $$

Conditions for Parallel and Perpendicular Lines

From the angle formula, we can derive conditions for lines to be parallel or perpendicular:

• Parallel Lines: Two lines with direction vectors $\vec{b_1}$ and $\vec{b_2}$ (or direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$) are parallel if the angle between them is $\theta = 0^\circ$. $\cos 0^\circ = 1$. This happens if and only if their direction vectors are collinear, i.e., one is a scalar multiple of the other: $\vec{b_1} = k \vec{b_2}$ for some non-zero scalar $k$.

  In terms of direction ratios, this means the direction ratios are proportional:

  $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

  (Handle zero denominators carefully: if $a_2=0$, then $a_1$ must be 0 for proportionality).
• Perpendicular Lines: Two lines with direction vectors $\vec{b_1}$ and $\vec{b_2}$ (or direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$) are perpendicular if the angle between them is $\theta = 90^\circ$. $\cos 90^\circ = 0$. This happens if and only if the dot product of their direction vectors is zero: $\vec{b_1} \cdot \vec{b_2} = 0$.

  In terms of direction ratios, this means the sum of the products of corresponding direction ratios is zero:

  $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$

  In terms of direction cosines: $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$.

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Answer:

Angle Between Two Planes

When two distinct planes in space are not parallel, they intersect to form a line. The angle between two intersecting planes is defined as the angle between the normal vectors to the planes. This angle is equal to the dihedral angle between the planes (or $180^\circ$ minus the dihedral angle, depending on which pair of normal vectors is chosen).

If the two planes are parallel, the angle between them is defined as $0^\circ$.

Similar to lines, the angle between two planes is conventionally taken as the acute angle between their normal vectors, i.e., between $0^\circ$ and $90^\circ$ ($\pi/2$).

Formula using Vector Form

Let the vector equations of two planes $P_1$ and $P_2$ be given by:

Here, $\vec{n_1}$ is a normal vector to plane $P_1$, and $\vec{n_2}$ is a normal vector to plane $P_2$. $d_1$ and $d_2$ are scalar constants related to the distance of the planes from the origin.

The angle $\theta$ between the planes $P_1$ and $P_2$ is defined as the angle between their normal vectors $\vec{n_1}$ and $\vec{n_2}$. Using the dot product formula for the angle between two vectors:

where $\phi$ is the angle between $\vec{n_1}$ and $\vec{n_2}$.

Since the angle between planes is taken as the acute angle $\theta$ ($0 \le \theta \le \frac{\pi}{2}$), we use the absolute value of the dot product:

$$ \mathbf{\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}} $$

where $0 \le \theta \le \frac{\pi}{2}$.

Formula using Cartesian Form

Let the Cartesian equations of two planes $P_1$ and $P_2$ be given by the general form:

The coefficients of $x, y, z$ in the general equation of a plane are the direction ratios of a normal vector to the plane.

A normal vector to plane $P_1$ is $\vec{n_1} = A_1\hat{i} + B_1\hat{j} + C_1\hat{k}$. The direction ratios are $(A_1, B_1, C_1)$.

A normal vector to plane $P_2$ is $\vec{n_2} = A_2\hat{i} + B_2\hat{j} + C_2\hat{k}$. The direction ratios are $(A_2, B_2, C_2)$.

The angle $\theta$ between the planes is the acute angle between $\vec{n_1}$ and $\vec{n_2}$. We use the formula $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.

Calculate the dot product $\vec{n_1} \cdot \vec{n_2}$ in component form:

Calculate the magnitudes $|\vec{n_1}|$ and $|\vec{n_2}|$:

Substitute these into the formula for $\cos \theta$:

$$ \mathbf{\cos \theta = \frac{|A_1 A_2 + B_1 B_2 + C_1 C_2|}{\sqrt{A_1^2 + B_1^2 + C_1^2} \sqrt{A_2^2 + B_2^2 + C_2^2}}} $$

This formula gives the cosine of the acute angle between two planes when their equations are given in Cartesian form.

Conditions for Parallel and Perpendicular Planes

From the angle formula, we can derive conditions for planes to be parallel or perpendicular:

• Parallel Planes: Two planes with normal vectors $\vec{n_1}$ and $\vec{n_2}$ (or coefficients $(A_1, B_1, C_1)$ and $(A_2, B_2, C_2)$) are parallel if the angle between them is $\theta = 0^\circ$. $\cos 0^\circ = 1$. This happens if and only if their normal vectors are collinear, i.e., one is a scalar multiple of the other: $\vec{n_1} = k \vec{n_2}$ for some non-zero scalar $k$.

  In terms of Cartesian coefficients, this means the coefficients of $x, y, z$ are proportional:

  $\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}$

  Note that for parallel planes, the constant terms $D_1$ and $D_2$ are generally different. If $\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} = \frac{D_1}{D_2}$, the planes are identical (coincident).
• Perpendicular Planes: Two planes with normal vectors $\vec{n_1}$ and $\vec{n_2}$ (or coefficients $(A_1, B_1, C_1)$ and $(A_2, B_2, C_2)$) are perpendicular if the angle between them is $\theta = 90^\circ$. $\cos 90^\circ = 0$. This happens if and only if their normal vectors are perpendicular: $\vec{n_1} \cdot \vec{n_2} = 0$.

  In terms of Cartesian coefficients, this means the sum of the products of corresponding coefficients of $x, y, z$ is zero:

  $A_1 A_2 + B_1 B_2 + C_1 C_2 = 0$

  This condition does not involve the constant terms $D_1$ and $D_2$.

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Answer:

Angle Between a Line and a Plane

When a straight line is not parallel to a plane and not contained within it, it intersects the plane at a single point. The angle between a line and a plane is defined geometrically as the angle between the line and its projection onto the plane.

Alternatively, a more convenient definition for calculation uses the normal vector of the plane. The angle between a line and a plane is defined as the complement of the angle between the line and the normal vector to the plane.

Let L be a line and P be a plane. Let $\vec{b}$ be the direction vector of the line L, and let $\vec{n}$ be a normal vector to the plane P.

Let $\theta$ be the angle between the direction vector of the line ($\vec{b}$) and the normal vector of the plane ($\vec{n}$).

The angle $\phi$ between the line L and the plane P is defined as:

From this relationship, taking the sine of both sides:

Using the trigonometric identity $\sin(\pi/2 - \theta) = \cos \theta$:

So, to find the sine of the angle between the line and the plane, we can calculate the cosine of the angle between the direction vector of the line and the normal vector of the plane. The angle $\theta$ between $\vec{b}$ and $\vec{n}$ is usually taken between $0$ and $\pi$, so $\cos\theta$ can be positive or negative. The angle $\phi$ between the line and the plane is usually taken as the acute angle ($0 \le \phi \le \pi/2$), for which $\sin\phi \ge 0$. Therefore, we use the absolute value of $\cos\theta$ in calculations involving $\sin\phi$.

Formula using Vector Form

Let the vector equation of the line be:

Here, $\vec{b}$ is the direction vector of the line L.

Let the vector equation of the plane be:

Here, $\vec{n}$ is a normal vector to the plane P.

Let $\phi$ be the angle between the line and the plane, and $\theta$ be the angle between the direction vector $\vec{b}$ and the normal vector $\vec{n}$. We know $\sin \phi = \cos \theta$.

The cosine of the angle $\theta$ between vectors $\vec{b}$ and $\vec{n}$ is given by the dot product formula:

To ensure $\sin \phi$ is non-negative for the acute angle $\phi$, we take the absolute value of the dot product:

Therefore, the sine of the angle $\phi$ between the line and the plane is:

$$ \mathbf{\sin \phi = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}} $$

Formula using Cartesian Form

Let the Cartesian equation of the line be:

Here, $(a, b, c)$ are the direction ratios of the line. A vector parallel to the line is $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$.

Let the Cartesian equation of the plane be:

Here, $(A, B, C)$ are the direction ratios of the normal vector to the plane. A normal vector to the plane is $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$.

Let $\phi$ be the angle between the line and the plane. Using the formula $\sin \phi = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$, we calculate the terms in component form:

Calculate the dot product $\vec{b} \cdot \vec{n}$:

Calculate the magnitudes $|\vec{b}|$ and $|\vec{n}|$:

Substitute these into the formula for $\sin \phi$:

$$ \mathbf{\sin \phi = \frac{|aA + bB + cC|}{\sqrt{a^2 + b^2 + c^2} \sqrt{A^2 + B^2 + C^2}}} $$

This formula gives the sine of the angle between a line and a plane when the direction ratios of the line and the normal vector of the plane are known.

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Answer:

Conditions for Parallelism and Perpendicularity of Lines and Planes

Based on the angle formulas derived for lines and planes, we can state clear conditions for when lines are parallel or perpendicular to each other, when planes are parallel or perpendicular to each other, and when a line is parallel or perpendicular to a plane. These conditions are essential for solving problems involving relative orientations.

1. Conditions for Parallelism and Perpendicularity of Two Lines

Let $L_1$ be a line with direction vector $\vec{b_1}$ (and direction ratios $a_1, b_1, c_1$). Let $L_2$ be a line with direction vector $\vec{b_2}$ (and direction ratios $a_2, b_2, c_2$).

• Condition for Perpendicular Lines ($L_1 \perp L_2$):

  Two lines are perpendicular if the angle between them is $90^\circ$. The angle between lines is the angle between their direction vectors. So, their direction vectors must be perpendicular. The dot product of perpendicular vectors is zero.

  Vector Condition: $\mathbf{\vec{b_1} \cdot \vec{b_2} = 0}$

  Cartesian Condition (using direction ratios): $\mathbf{a_1 a_2 + b_1 b_2 + c_1 c_2 = 0}$

• Condition for Parallel Lines ($L_1 \parallel L_2$):

  Two lines are parallel if the angle between them is $0^\circ$. This occurs if and only if their direction vectors are collinear. One vector must be a non-zero scalar multiple of the other.

  Vector Condition: $\mathbf{\vec{b_1} = \lambda \vec{b_2}}$ for some scalar $\lambda \neq 0$.

  Cartesian Condition (using direction ratios): The corresponding direction ratios must be proportional.

  $\mathbf{\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}}$

  (assuming $a_2, b_2, c_2 \neq 0$)

  If a denominator is zero (e.g., $a_2=0$), the corresponding numerator must also be zero ($a_1=0$) for the lines to be parallel.

2. Conditions for Parallelism and Perpendicularity of Two Planes

Let $P_1$ be a plane with normal vector $\vec{n_1}$ (and normal direction ratios $A_1, B_1, C_1$). Let $P_2$ be a plane with normal vector $\vec{n_2}$ (and normal direction ratios $A_2, B_2, C_2$). The angle between planes is the angle between their normal vectors.

• Condition for Perpendicular Planes ($P_1 \perp P_2$):

  Two planes are perpendicular if the angle between them is $90^\circ$. This means their normal vectors must be perpendicular. The dot product of their normal vectors is zero.

  Vector Condition: $\mathbf{\vec{n_1} \cdot \vec{n_2} = 0}$

  Cartesian Condition (using normal direction ratios): $\mathbf{A_1 A_2 + B_1 B_2 + C_1 C_2 = 0}$

• Condition for Parallel Planes ($P_1 \parallel P_2$):

  Two planes are parallel if the angle between them is $0^\circ$. This occurs if and only if their normal vectors are collinear. One normal vector must be a non-zero scalar multiple of the other.

  Vector Condition: $\mathbf{\vec{n_1} = \lambda \vec{n_2}}$ for some scalar $\lambda \neq 0$.

  Cartesian Condition (using normal direction ratios): The corresponding coefficients of $x, y, z$ must be proportional.

  $\mathbf{\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}}$

  (assuming $A_2, B_2, C_2 \neq 0$)

  If these ratios are also equal to $\frac{D_1}{D_2}$, then the planes are identical (coincident).

3. Conditions for Parallelism and Perpendicularity of a Line and a Plane

Let line L have direction vector $\vec{b}$ (and direction ratios $a, b, c$). Let plane P have normal vector $\vec{n}$ (and normal direction ratios $A, B, C$). Let $\phi$ be the angle between the line and the plane, and $\theta$ be the angle between $\vec{b}$ and $\vec{n}$. Recall that $\phi = 90^\circ - \theta$.

• Condition for a Line Perpendicular to a Plane (Line $\perp$ Plane):

  A line is perpendicular to a plane if the angle between them is $90^\circ$, i.e., $\phi = 90^\circ$. This means $\theta = 90^\circ - 90^\circ = 0^\circ$. So, the angle between the line's direction vector $\vec{b}$ and the plane's normal vector $\vec{n}$ is $0^\circ$. This occurs if and only if the line's direction vector is parallel to the plane's normal vector.

  Vector Condition: $\mathbf{\vec{b} = \lambda \vec{n}}$ for some scalar $\lambda \neq 0$. (Direction vector of line is parallel to normal vector of plane).

  Cartesian Condition: The direction ratios of the line must be proportional to the direction ratios of the normal vector.

  $\mathbf{\frac{a}{A} = \frac{b}{B} = \frac{c}{C}}$

  (assuming $A, B, C \neq 0$)

• Condition for a Line Parallel to a Plane (Line $\parallel$ Plane):

  A line is parallel to a plane if the angle between them is $0^\circ$, i.e., $\phi = 0^\circ$. This means $\theta = 90^\circ - 0^\circ = 90^\circ$. So, the angle between the line's direction vector $\vec{b}$ and the plane's normal vector $\vec{n}$ is $90^\circ$. This occurs if and only if the line's direction vector is perpendicular to the plane's normal vector. The dot product of perpendicular vectors is zero.

  Vector Condition: $\mathbf{\vec{b} \cdot \vec{n} = 0}$. (Direction vector of line is perpendicular to normal vector of plane).

  Cartesian Condition: The sum of the products of corresponding direction ratios must be zero.

  $\mathbf{aA + bB + cC = 0}$

  Note that this condition $aA + bB + cC = 0$ only guarantees that the line is parallel to the plane or lies within the plane. To check if it lies within the plane, verify if any point on the line (e.g., $(x_1, y_1, z_1)$ from the line equation) also satisfies the plane equation $Ax_1 + By_1 + Cz_1 + D = 0$. If it satisfies the equation, the line lies in the plane. If it does not satisfy the equation, the line is strictly parallel to the plane.

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